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In the second-order approach to Peano Arithmetic, the only non-logical symbols are the constant $0$ and the successor function $S(*).$ But, when we go to first-order Peano Arithmetic, something goes wrong with this approach, and we need to include addition and multiplication among our non-logical symbols, as well as a whole slew of axioms in this larger language.

What goes wrong?

For instance, I've read that we can define addition in second-order arithmetic by writing

  • $x+0 = x$
  • $x+S(y) = S(x+y).$

Why does this work in second order arithmetic, but not in first-order?

goblin GONE
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  • See http://web.mat.bham.ac.uk/R.W.Kaye/logic/succ – Carl Mummert Jul 22 '13 at 02:05
  • I added the tag "logic" which is likely to have many more followers. I think it's a good idea in general to tag most logic questions with that tag and additional more-specific tags – Carl Mummert Jul 22 '13 at 02:10
  • @CarlMummert, thanks, I'll keep that in mind. – goblin GONE Jul 22 '13 at 02:14
  • @CarlMummert, I've read over that link a few times, but I'm not quite sure of the significance. Are you saying that, 'okay, we COULD leave $+$ and $\times$ out of our non-logical symbols, but this would make quantifier elimination harder to prove'? – goblin GONE Jul 22 '13 at 14:59
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    Peano arithmetic does not have quantifier elimination. Quantifier elimination, like decidability, is a sign that a foundational theory is not very strong. – Carl Mummert Jul 23 '13 at 01:01

3 Answers3

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There is a point of confusion in the question:

For instance, I've read that we can define addition in second-order arithmetic by writing

  • $x+0 = x$
  • $x+S(y) = S(x+y).$

Why does this work in second order arithmetic, but not in first-order?

That does not work in second-order arithmetic. It is an implicit characterization of the addition function, but it is not an explicit definition of the addition function in terms of the successor function.

A genuine definition is a formula $\phi(n,m,p)$ in a language without the addition symbol $+$ such that, for all natural numbers $n,m,p$, we have $n + m = p$ if and only if $\phi(n,m,p)$ holds. A "pseudo-definition" that is able to refer to the object being defined is called an implicit definition, but implicit definability is much weaker than actual definability.

One actual definition of addition of natural numbers in second-order arithmetic is: $$ n + m = p \Leftrightarrow (\forall f)\left[ \left( f(0) = n \land (\forall k)[f(S(k)) = S(f(k)]\right ) \to f(m) = p\right]. $$ Here $n,m,p$ are natural numbers, $(\forall k)$ quantifies over the natural numbers, and $(\forall f)$ quantifies over all unary functions from the natural numbers to themselves. Notice that, crucially, the right side does not mention $+$. In the particular definition, we could also rewrite it with an existential function quantifier: $$ n + m = p \Leftrightarrow (\exists f)\left( f(0) = n \land (\forall k)[f(S(k)) = S(f(k))] \land f(m) = p\right.) $$

Why does this definition not work in first-order logic? Because, in a single-sorted first-order theory of arithmetic, it is not possible to quantify over functions in the way that the definition does.

Now, that does not prove that it is impossible to define addition of natural numbers in terms of successor. It only shows that the definition in second-order arithmetic does not go through unchanged.

One way to see that addition is not definable from successor is sketched in this answer by Alex Kruckman. The key point is that if we look at the first-order theory of the natural numbers with successor and a constant for 0, every formula in this language (with some free variables) is equivalent to a quantifier-free formula in the language (with the same free variables). A proof of that is given by Richard Kaye here. So if addition was definable in that structure, it would be definable by a quantifier-free formula. But by analyzing the form of such a formula we can show that it cannot define addition.

Actually, more is known. Neither addition nor multiplication is definable from successor alone; multiplication is not definable from successor and addition; but addition is definable over $\mathbb{N}$ from successor and multiplication. The theory of the natural numbers with multiplication and addition is undecidable, but the restriction to just addition is decidable, and the restriction with just multiplication is decidable. The reference A Survey of Arithmetical Definability by Bés was mentioned in the comments, with much more detailed results.

Carl Mummert
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    Hm. Interesting. Since successorship is definable from the order (on $\Bbb N$, that is), we can define $0,S,+,\cdot$ and show that the axioms of $\sf PA_2$ hold, just in the second-order structure $\langle\Bbb N,\leq\rangle$. That's very nice. (And now that I've written that, it makes a lot of sense because we actually define all those in the language of set theory where we only have $\in$ which functions as the surrogate for $<$...) – Asaf Karagila Jul 23 '13 at 01:05
  • @Asaf: this is really all a consequence of the fact that $\omega$ is the smallest (positive) limit ordinal, and as such has a very particular order structure that can be leveraged to make a model of PA in settings like second-order logic and ZFC. You can actually define a model of PA in any full second-order structure with an infinite domain and any language that includes equality. – Carl Mummert Jul 23 '13 at 01:09
  • Ah. I see why (although you should require the structure is countable, not just infinite). We can say when a set is finite, so we can define a linear order that every initial segment is finite. From that we can finish the rest. – Asaf Karagila Jul 23 '13 at 01:14
  • @Asaf: since you can define countability in SOL and quantify over arbitrary subsets, you can always pull out a countable subset of an infinite domain. So you can get a model of PA in the sense that you can prove "there is a countable set $N$, an element $0 \in N$, and operations $S, +, \times \colon N \to N$ such that $(N,S,+,\times,0, =)$ satisfies Peano arithmetic". – Carl Mummert Jul 23 '13 at 01:17
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    +5 for @CarlMummert's original answer. Surely a model of clarity and helpfulness! – Peter Smith Jul 23 '13 at 09:45
  • @PeterSmith, agreed! This is an exceptional answer. – goblin GONE Jul 23 '13 at 20:03
  • I don't see anything wrong with the recursive definition as given (assuming suitable quantifiers). Using the axioms of set theory, you can construct a unique function (a set of ordered triples of natural numbers) with just that property, i.e. you prove the existence of such a function on the natural numbers. – Dan Christensen Jul 25 '13 at 05:15
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    @Dan Christensen: the issue is that neither ZFC nor second-order arithmetic has any syntax that literally makes the two recursion equations into a "definition". It is true that for well-posed recursive definitions there are techniques for converting the recursion equations into an actual definition; I indicate one in the answer. But the question claimed that we can define addition simply "by writing" the equations, when there is actually much more going on. It is easy to write a system of recursion equations that doesn't define any function. – Carl Mummert Jul 25 '13 at 10:58
  • @CarlMummert Yes, some set-theoretic rationale must be presented for the recursive definition given. But, having established this rationale, the above can be used as a workable definition that completely characterizes the addition function, can it not? – Dan Christensen Jul 25 '13 at 16:53
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    Sorry for digging out the old topic, but I have a question: can we do something similar in the language of second-order arithmetic which allows quantification over sets, and not functions? – Wojowu Oct 23 '14 at 19:35
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    @DanChristensen The above recursive formulation does indeed characterize addition, but its not a definition. In a true definition you can replace all occurrences of the function by the right hand side. Since you can't do that using the given characterization, you have to include those operators in the signature in order to use them and constrain them with axioms, and that is indeed what Peano does: (N,0,s,+,.,<). But the price you pay is that the theory of this structure is much richer than the theory of (N,0,s,<) (which has a decision procedure via Q.E.), and it turns out it is incomplete. – Motorhead Dec 04 '15 at 20:54
  • @S.N. How about: $$\forall \space a,b,c\space [a+b=c \iff (a,b,c)\in N^3$$ $$ \land \forall d\subset N^3\space [\forall e\in N: (e,0,e)\in d $$ $$\land \forall e,f,g \space [(e,f,g)\in d \implies (e,S(f),S(g))\in d] \implies (a,b,c)\in d]]?$$ – Dan Christensen Dec 05 '15 at 03:41
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    @DanChristensen You‘re quantifying over sets which is no longer 1st order – Motorhead Dec 06 '15 at 08:10
  • @S.N. To my knowledge, most mathematicians don't feel constrained to first-order theories. In any case, it is a "true definition" in your sense that "you can replace all occurrences of the function by the right hand side." – Dan Christensen Dec 06 '15 at 08:57
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    @DanChristensen: Certainly. But the title of this post is "Why are addition and multiplication included in the signature of first-order Peano arithmetic?". If you're interested in 2nd order, Carl Mummert explains how to define + in SOL. – Motorhead Dec 07 '15 at 19:07
  • @S.N. I was addressing your concern that you wanted the definition in a non-recursive form. I'm never quite sure what is meant by FO and SO in these contexts. I will stick to applications of "standard" set theory where I can construct arithmetic functions from the "usual" five Peano axioms. The precise construction of the addition function is more obvious in this definition. It's just a subset of $N^3$. – Dan Christensen Dec 07 '15 at 20:57
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    @DanChristensen I don't recall asking for anything :-) I was simply trying to help out with this question of yours: "But, having established this rationale, the above can be used as a workable definition that completely characterizes the addition function, can it not?" I hope that my answer helped. – Motorhead Dec 09 '15 at 22:07
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    I just stumbled on this old answer and wanted to make a quick correction. In the last paragraph, you say "addition is not definable from successor and multiplication". Actually, $a+b=c$ is definable by $S(a\cdot c)\cdot S(b\cdot c) = S(c\cdot c \cdot S(a\cdot b))$ (this works when $c\neq 0$, and we can handle the case $c = 0$ separately). This clever definition is due to Julia Robinson from her paper "Decidability and decision problems in arithmetic". – Alex Kruckman Nov 23 '20 at 22:28
  • @AlexKruckman, this is called Tarski's identity, after Alfred Tarski, a Polish mathematician. – ByteEater Aug 11 '22 at 09:11
  • @ByteEater Thanks, I had not known that the identity had a name. Of course I know who Alfred Tarski is! In addition to being a Polish mathematician, he was one of the progenitors of the field of model theory, the founder of the logic group at Berkeley, and Julia Robinson's PhD thesis advisor. – Alex Kruckman Aug 11 '22 at 12:32
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    @ByteEater I'm a bit confused as to why the identity should be called "Tarski's" rather than "Robinson's" though. The identity makes it obvious that addition is definable from multiplication and successor. Tarski was apparently very impressed when Robinson solved this problem (see Feferman's short biography of Robinson, top of page 7). This seems to contradict the idea that Tarski knew this identity before Robinson did. – Alex Kruckman Aug 11 '22 at 12:37
  • Indeed, nice reference! Having read p. 7, I concur that it's unlikely he knew it before. Perhaps Kephir can clarify the confusion, the original author of Skolem arithmetic Wikipedia article. – ByteEater Aug 11 '22 at 13:31
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    @ByteEater The reference on Wikipedia no doubt originated with the cited article A Survey of Arithmetical Definability by Bés, which on p. 32 credits the result to Robinson but calls the identity Tarski's. Odd! – Alex Kruckman Aug 11 '22 at 14:38
  • @Alex Kruckman - thank you for pointing that out, I edited the answer. I remember looking at some classic paper for reference when I wrote this answer, but I can't remember what paper that was.... I wonder if we are talking about two different things: definability over $\mathbb{N}$ in various languages vs. definabililty over fragments of PA. Do you know if PA is interpretable in the restriction of PA to the language with only $\times$ and $S$? The revised answer is definitely correct from J. Robinson's work, in any case. – Carl Mummert Aug 12 '22 at 11:45
  • I should be able to check if Robinson's identity works to give that interpretation, of course. But, although I might be wrong, I thought the answer was negative for some reason when I wrote the answer originally. – Carl Mummert Aug 12 '22 at 12:14
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The reason is that it is impossible to define addition and multiplication using only successor operation in first-order logic.

Godot
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    For example, if we weaken Peano arithmetic by removing the multiplication sign and any axiom that mentioned it, we obtain Presberger arithmetic, which is known to be a much weaker theory than Peano arithmetic. If multiplication was definable in Presberger arithmetic, that could not be the case. – Carl Mummert Jul 22 '13 at 01:58
  • @Godot, why does the definition $x+0 = x$, $x+S(y)=S(x+y)$ suffice for defining addition in second-order arithmetic, but not in first order? – goblin GONE Jul 22 '13 at 15:00
  • @CarlMummert Weaker in what sense? Presberger arithmetic is decideable, but Peano arithmetic isn't. Doesn't deciability count for anything in terms of strength? – Doug Spoonwood Jul 23 '13 at 00:18
  • @Doug Spoonwood: any sufficiently strong foundational theory is undecidable, and in fact essentially undecidable. Decidability of a theory is evidence that they theory is not very strong at all. – Carl Mummert Jul 23 '13 at 00:31
  • @user18921 Please check Carl's answer. I don't think that I can add much. – Godot Jul 23 '13 at 12:30
  • @CarlMummert So what does strength mean? Does it mean anything at all? As I understand things propositional calculi comes as more foundational than anything else. Propositional calculi, at least usually if not always, are decideable. But what you've said implies the very foundations of logic as not strong at all. Do you really mean to imply that? – Doug Spoonwood Jul 23 '13 at 23:26
  • Downvoting because this answer doesn't say more than the question implies. – Arvid Samuelsson Dec 09 '21 at 12:46
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In a 2nd-order approach to PA, using the axioms of set theory, it is possible to construct the usual addition and multiplication (and exponentiation) functions starting with only a $0$ (or $1$) and a successor function.

In a 1st-order approach, since set theory is not available, you will need extra axioms for the addition and multiplication (and exponentiation) functions.

Dan Christensen
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